AMC 12 Mock Exam 4 Problem 19

Problem 19 of 25

Regular hexagon $ABCDEF$ has side length $2$ . A point $P$ inside the hexagon is chosen so that the area of $\triangle PAB$ is $1$ and the area of $\triangle PCD$ is $2$ , as shown (the figure is not necessarily drawn to scale). What is the area of $\triangle PEF$ ?

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Solution

Extend sides $\overline{AB}$ , $\overline{CD}$ , and $\overline{EF}$ . Their lines bound an equilateral triangle $T$ of side $3\cdot2=6$ that contains the hexagon (the hexagon is $T$ with three corner triangles of side $2$ removed). For a point $P$ inside $T$ , the sum of its distances to the three sides of $T$ equals the height of $T$ , which is $3\sqrt3$ . (Split $T$ into three triangles with apex $P$ : $\frac12\cdot6\,(d_1+d_2+d_3)=[T]=9\sqrt3$ .)

Since $AB=CD=EF=2$ , the areas $[PAB]$ , $[PCD]$ , $[PEF]$ equal $\frac12\cdot2\cdot d=d$ for the corresponding distances. Hence $[PAB]+[PCD]+[PEF]=3\sqrt3$ , and $[PEF]=3\sqrt3-1-2=\boxed{3\sqrt3-3}$ . (This equals half the hexagon's area, $6\sqrt3/2$ , minus $3$ .)

Tagged: regular hexagon, areas from an interior point (Viviani)

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